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Force per unit length between two long straight parallel conductors. Suppose two thin straight conductors (or wires) PQ and RS are placed parallel to each other in vacuum (or air) carrying current \[{I_1}\;and\;{I_2}\] respectively. It has been observed experimentally that when the currents in the wire in the same direction, they experience an attractive force and when they carry current in opposite directions, they experience a repulsive force. Let the conductors PQ and RS carry currents \[{I_1}\;and\;{I_2}\] in the same direction and placed at separation r.

\[{B_1} = \dfrac{{{\mu _0}{I_1}}}{{2\pi r}}\]

\[\Delta F = {B_1}{I_1}\;\Delta L\;\sin \;90^\circ \]

\[{B_1} = \dfrac{{{\mu _0}{I_1}}}{{2\pi r}}{I_2}\;\Delta L\]

\[F = \dfrac{{{\mu _0}{I_1}{I_2}}}{{2\pi r}}\sum {\Delta L} = \dfrac{{{\mu _0}{I_1}{I_2}}}{{2\pi r}}L\]

\[\dfrac{F}{L} = \dfrac{{{\mu _0}{I_1}{I_2}}}{{2\pi r}}N/m\]

Consider a current-element â€˜abâ€™ of length \[\Delta L\] of wire RS. The magnetic field produced by current carrying conductor PQ at the location of other wire RS.

According to Maxwellâ€™s right hand rule or right hand Palm rule no. 1, the direction of B

\[\dfrac{F}{L} = \dfrac{{{\mu _0}{I_1}{I_2}}}{{2\pi R}}N/m\]

\[F = \dfrac{{{\mu _0}}}{{2\pi }} = 2 \times {10^{ - 7}}N/m\]

Thus 1 ampere is the current which when flowing in each of parallel conductors placed at separation 1 m in vacuum exert a force of \[2 \times {10^{ - 7}}\] on 1m length of either wire.

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